Data is not inserting into table? - php

Question

GYZ i dont know why data is not inserting in my data base #Mysql
. infact im using mysqli_connect and mysql_connect both ,I'm still facing same prob ..this is my code:
<?php
$servername = 'localhost';
$username = 'root';
$password = '';
$db='school';
#mysqli_connect($servername, $username, $password);
#mysqli_select_db($db); //or die ('not Connect to db ');
if(isset($_GET['submit'])) {
$sid= $_GET['sid'];
$sname= $_GET['sname'];
$fname= $_GET['fname'];
$order= #mysqli_query("insert into school (sid,sname,fname) values ('$sid','$sname','$fname');");
if ($order) {
echo '<br>Input data is successful';
} else {
echo '<br>Input data is not valid';
}
} ?>

Answer 1

I revisited the question and posted the following, seeing that nobody posted one.
You didn't pass the db connection to mysqli_select_db() nor for mysqli_query() and need to assign a variable to the connection first.
Both of those require it in mysqli_ and you may have been accustomed to mysql_ in the past. MySQLi_ is different than MySQL_ when it comes to certain functions that needs a connection.
Sidenote: The # symbol is an error suppressor. Remove it during testing/development.
Another sidenote: Both your database and table bear the same name of school. Make sure that this is correct.
<?php
$servername = 'localhost';
$username = 'root';
$password = '';
$db='school';
$connect = mysqli_connect($servername, $username, $password, $db);
if($connect){
echo "Connected";
}
else {
echo "Error: " . mysqli_error($connect);
}
// This isn't needed. You can pass all 4 parameters in one shot.
// $database = mysqli_select_db($connect, $db); //or die ('not Connect to db ');
if(isset($_GET['submit'])) {
$sid= $_GET['sid'];
$sname= $_GET['sname'];
$fname= $_GET['fname'];
$order= mysqli_query($connect, "INSERT INTO school (sid,sname,fname) VALUES ('$sid','$sname','$fname');");
if ($order) {
echo '<br>Input data is successful.';
} else {
// Uncomment the one below once everything is ok.
// echo '<br>Input data is not valid.';
// Comment this below once there are no errors.
echo "There was an error: " . mysqli_error($connect);
}
}
References:
http://php.net/manual/en/function.mysqli-connect.php
http://php.net/manual/en/mysqli.select-db.php
http://php.net/manual/en/mysqli.query.php
Check for errors also via PHP and the query:
http://php.net/manual/en/mysqli.error.php
http://php.net/manual/en/function.error-reporting.php
And make sure you're running this off a webserver, or if local that PHP/MySQL are installed, running properly and using http://localhost as opposed to file:///.
Your code is also open to an SQL injection, use a prepared statement.
References:
https://en.wikipedia.org/wiki/SQL_injection
https://en.wikipedia.org/wiki/Prepared_statement
Footnotes:
You seem to want to use this in a table. <form> cannot be child of <table> if you are using those tags outside of the form which wasn't posted in your question; there are stray <td></td> tags.